Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{4}{3(z - 5)} \times \dfrac{5(z - 5)}{2z} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 4 \times 5(z - 5) } { 3(z - 5) \times 2z } $ $ a = \dfrac{20(z - 5)}{6z(z - 5)} $ We can cancel the $z - 5$ so long as $z - 5 \neq 0$ Therefore $z \neq 5$ $a = \dfrac{20 \cancel{(z - 5})}{6z \cancel{(z - 5)}} = \dfrac{20}{6z} = \dfrac{10}{3z} $